Strange Capacitors...
Philldapill
Posts: 1,283
A couple of semesters back in Electromagnetics class, I got to thinking about capacitors. The "thinking" goes like this.
Say you have a custom made, square plate,·capacitor with plate spacing of X, and area A. This capacitor is made in such a way that it has about 98% liquid water, and 2% water vapor between the plates. Thos numbers aren't that important, but the idea is that there is NO air between the plates, and the volume is perfectly sealed with no ability to expand/contract.
What you have now, is a normal square plate capacitor with pure water as the dielectric. Pretend this is a perfect capacitor with no leakage. Now charge the capacitor to, say, 10V. The charge on the plates is: q = CV, so it's a function of the capacitance and the voltage. Energy stored in the capacitor is 1/2*C*V^2.
Now, FREEZE the water between the plates. The water will expand slightly, but because we had the small bit of volume with water vapor in it, the ice can expand between the plates, but the SAME amount of water molecules are between the plates. Since the dielectric constant of ice is less than that of water, the capacitance should change. Because the same amount of charge is still on the plates, the voltage should increase. With that increase in voltage, the energy on the plates went up.
WHAT? Where did this energy come from? Looking at this whole thing from a thermodynamic standpoint, we initially added energy to the system via charging the plates. We then REMOVED heat from the system(freezing). At the end, we now have MORE energy in the system than we should? What in the heck is going on here?
Anyone have a clue as to where the flaw is? If you're guessing my assumption about the dielectric of ice being different than water, my prof and I are 99.9% sure of that being true. Heck, even wikipedia says it's true...
EDIT: Just another side note - One theory is that the freezing point of water is lower when in an electric field. The reasoning is that the molecules are already highly polarized to the field, and resist lining up in a crystal structure to form ice. Even if that's the case, you are still pulling heat from the system(just more of it), and gaining energy on the plates once the water freezes.
Post Edited (Philldapill) : 7/22/2009 7:26:26 PM GMT
Say you have a custom made, square plate,·capacitor with plate spacing of X, and area A. This capacitor is made in such a way that it has about 98% liquid water, and 2% water vapor between the plates. Thos numbers aren't that important, but the idea is that there is NO air between the plates, and the volume is perfectly sealed with no ability to expand/contract.
What you have now, is a normal square plate capacitor with pure water as the dielectric. Pretend this is a perfect capacitor with no leakage. Now charge the capacitor to, say, 10V. The charge on the plates is: q = CV, so it's a function of the capacitance and the voltage. Energy stored in the capacitor is 1/2*C*V^2.
Now, FREEZE the water between the plates. The water will expand slightly, but because we had the small bit of volume with water vapor in it, the ice can expand between the plates, but the SAME amount of water molecules are between the plates. Since the dielectric constant of ice is less than that of water, the capacitance should change. Because the same amount of charge is still on the plates, the voltage should increase. With that increase in voltage, the energy on the plates went up.
WHAT? Where did this energy come from? Looking at this whole thing from a thermodynamic standpoint, we initially added energy to the system via charging the plates. We then REMOVED heat from the system(freezing). At the end, we now have MORE energy in the system than we should? What in the heck is going on here?
Anyone have a clue as to where the flaw is? If you're guessing my assumption about the dielectric of ice being different than water, my prof and I are 99.9% sure of that being true. Heck, even wikipedia says it's true...
EDIT: Just another side note - One theory is that the freezing point of water is lower when in an electric field. The reasoning is that the molecules are already highly polarized to the field, and resist lining up in a crystal structure to form ice. Even if that's the case, you are still pulling heat from the system(just more of it), and gaining energy on the plates once the water freezes.
Post Edited (Philldapill) : 7/22/2009 7:26:26 PM GMT
Comments
If you run this experiment in a closed-loop calorimeter, you'd see what's missing in the thought process.
That's the hint ;-P
- H
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The analysis doesn't depend at all on the kind of capacitor.· It's the same with any variable capacitance.· With constant charge Q. reducing the capacitance C must incease the voltage E [noparse][[/noparse]electromotive force; engineers and physicists don't often·use V for voltage] -- and, from conservation of energy,·the changes to C and E must be in the ratio that maintains the energy e=1/2 CE2 at a constant value.·
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· -- Carl, nn5i@arrl.net
In another example, we could have a removable dielectric in the capacitor. If you try to pull the dielectric out, it will require some force and by doing so, requires energy. In that case, there is an energy balance in the system - energy in(pulling the dielectric out) = energy out(increase in capacitor energy as capacitance drops).
Where the heck have you been? We've missed you around here!
-Phil
Q_after = Q_before = Q --- conservation of charge
C_after = C_before / 80 --- effect of draining water on capacitance
Q = C * V
===>
V_after = V_before * 80
But what about energy, U?
U = C * V2 / 2
U_after = (C_before / 80) * (V_before * 80)2 / 2
. = U_before * 80
How can that be? Perpetual motion? A hidden fountain of energy with only a capacitor and a pump?
The resolution is, I think, that work is done to remove the water from between the plates. Strange, huh? The water molecules are polarized by the electric field, lined up more or less +-+-+- across the dielectric layer, and the work necessary to drain the water ends up back on the capacitor. It would take work to refill the capacitor with water, and that voltage would drop back down and the energy would be respent. No perpetual motion.
In the case of freezing, the water has fewer degrees of freedom in the crystal lattice. The capacitance goes down, and the voltage and energy go up. It does seem reasonable that there would be a change in the freezing point that could be detected in a sensitive experiment.
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Tracy Allen
www.emesystems.com
Let's say, initially, the capacitance is C1=2 and the voltage is E1=1.· Charge Q=C1E1=2, and energy e=1/2C1E12=1.
Reduce the capacitance by half.· New capacitance C2=1, new voltage·E2=2, Q=C2E2=2, e=1/2C2E22=2.
The charge Q remained constant at 2, and the energy e increased, just as you say, and just as I missed.·
Clearly it required an energy input to decrease the capacitance while maintaining constant charge.· If, for example, you decreased the capacitance by moving the plates apart, you did some work in order to move them apart against the attraction of their opposite charges.
If you changed the capacitance in some other way, such as freezing the dielectric (water), clearly some of the energy from the heat of solidificatiion must have been used in reducing the capacitance, and this would diminish the heat you·extract by freezing the water.· We both failed to understand that, but at least you realized you were missing something.· I didn't realize I was.
Incidentally, I prefer U for energy now that I read Tracy's post.
Phil, I've been busy with other things.
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· -- Carl, nn5i@arrl.net
Post Edited (Carl Hayes) : 7/23/2009 12:08:26 AM GMT
So, it looks like we are all sort of in agreement that by applying an electric field to the water molecules, the freezing/melting point would probably be reduced? If this is the case, let's take the E field to the extreme... Increase the E field right up to the point of the dielectric breakdown of water. If the freezing point is reduced enough(>10 degrees F), we could lower the temperature right above the "new" freezing point, remove the E field, and look at the water. What does it do now? Does it instantly freeze solid?
If the water DOES freeze solid extremely quickly(kind of like superheated water instantly boiling it when you bump it), couldn't this process be used in things like flash freezing sensitive foods?
Just for the record... I'm fully in the "Perpetual Motion is BS" camp. I knew I was missing something, and my questions are more aimed at finding what it is I'm missing, not claiming I've discovered zero-point energy. [noparse]:)[/noparse]
On another note, there is an unrelated thread of someone calling people on this forum, er, not nice things... This is the only forum I know of where this sort of discussion happens. Thank you guys!
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propmod_us and propmod_1x1 are in stock. Only $30. PCB available for $5
Want to make projects and have Gadget Gangster sell them for you? propmod-us_ps_sd and propmod-1x1 are now available for use in your Gadget Gangster Projects.
Need to upload large images or movies for use in the forum. you can do so at uploader.propmodule.com for free.